Question

The molecular weight of KMnO4 is 158 g/mol. If you wanted to make 500 mL of a 0.1M stock solution of KMnO4, how much solid KMnO4 would you need to add before filling mL flask of water?

Answer 1

Solid KMnO₄ needed = 7.9 g

Further explanation

Given

MW KMnO₄ = 158 g/mol

500 mL(0.5 L) of a 0.1M stock solution of KMnO₄

Required

solid KMnO₄

Solution

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

$\large{\boxed {\bold {M ~ = ~ \frac {n} {V}}}$

Input the value :

n = M x V

n = 0.1 M x 0.5 L

n = 0.05 mol

Mass KMnO₄ :

= mol x MW

= 0.05 x 158 g/mol

= 7.9 g